Enthalpy of Formation: Solve CO + O_2 to CO_2

• ubiquinone
In summary, the question relates to the calculation of heat released during combustion of CO(g) in excess oxygen. The molar enthalpies of CO, O_2, and CO_2 are provided for the equation. The solution involves finding the change of enthalpy and multiplying it by the number of moles of CO.
ubiquinone
Hi, I have a question involving enthalphy of formation. I tried doing this problem but I couldn't seem to get the right answer. I was wondering if anyone here may please help me out. Thanks.

Question: How much heat is evolved when 13.39 moles of CO(g) is burned in excess oxygen.
$$2CO(g) + O_2(g)\rightarrow 2CO_2(g)$$
Molar enthalpies are
$$CO=-110.5 kJ/mol$$
$$O_2=0.0$$
$$CO_2=-393.5 kJ/mol$$

you figure out the Change of Enthalpy (change of H) of overall equation, and then multiply by 13.39 moles of CO to get energy (KJ).

just to make sure...

Change of H = [(2)(-110.5KJ)] -[(2)(-110.5KJ)] = ?? KJ

then mult ?? KJ by 13.39 mol

does it give you the right answer?? would like to know

To solve this problem, we need to use the concept of enthalpy of formation, which is the change in enthalpy when a compound is formed from its constituent elements in their standard states. In this case, we are given the molar enthalpies of formation for CO, O_2, and CO_2.

First, we need to write the balanced chemical equation for the reaction:
2CO(g) + O_2(g) → 2CO_2(g)

Next, we need to determine the molar enthalpy change for the reaction. This can be done by using the following formula:
ΔH = ∑nΔHf(products) - ∑nΔHf(reactants)
Where ΔH is the change in enthalpy, n is the number of moles, and ΔHf is the molar enthalpy of formation.

In this case, we have 13.39 moles of CO reacting, so we will use the molar enthalpy of formation for CO. We also have 1 mole of O_2 reacting, so we will use the molar enthalpy of formation for O_2. And finally, we have 13.39 moles of CO_2 being formed, so we will use the molar enthalpy of formation for CO_2.

Plugging in the values, we get:
ΔH = (13.39 mol)(-393.5 kJ/mol) - (13.39 mol)(-110.5 kJ/mol) - (1 mol)(0.0 kJ/mol)
ΔH = -5260.2 kJ/mol

This means that for every 2 moles of CO that react, 5260.2 kJ of heat is released. Since we have 13.39 moles of CO, we need to divide the value by 2 and then multiply by 13.39 to get the total heat released:
(5260.2 kJ/mol ÷ 2 mol) x 13.39 mol = 35271.6 kJ

Therefore, the heat evolved when 13.39 moles of CO is burned in excess oxygen is 35271.6 kJ. I hope this helps! If you are still having trouble, feel free to ask for further clarification.

What is enthalpy of formation?

Enthalpy of formation is the amount of heat released or absorbed when one mole of a compound is formed from its constituent elements in their standard states.

Why is it important to calculate enthalpy of formation?

Enthalpy of formation is important because it helps us understand the stability and reactivity of a compound. It also allows us to predict the amount of energy released or required in a chemical reaction.

How do you calculate enthalpy of formation?

Enthalpy of formation can be calculated using the following formula: ΔHf = ΣnΔHf(products) - ΣmΔHf(reactants), where n and m are the coefficients of the products and reactants respectively.

What are the units of enthalpy of formation?

The units of enthalpy of formation are kilojoules per mole (kJ/mol) or kilocalories per mole (kcal/mol).

Can enthalpy of formation be negative?

Yes, enthalpy of formation can be negative, indicating that the formation of the compound releases heat (exothermic reaction). It can also be positive, indicating that the formation of the compound requires heat (endothermic reaction).

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